Okay, so the problem is asking for the least positive integer multiple of 30 that can be written using only the digits 0 and 2. Let me start by understanding what exactly is being asked here. 

First, a multiple of 30 means the number must be divisible by 30. Now, 30 factors into 2 * 3 * 5, so the number must be divisible by 2, 3, and 5. Let me break down the divisibility rules for each of these:

1. Divisible by 2: The last digit must be even. Since the digits allowed are only 0 and 2, this is automatically satisfied because both 0 and 2 are even. So, the number can end with 0 or 2. But since we're looking for the least such number, ending with 0 might be better because that uses fewer digits, but I need to check.

2. Divisible by 3: The sum of the digits must be divisible by 3. Since the digits are only 0s and 2s, the sum will be a multiple of 2 (because each 2 contributes 2). So, the sum of the digits must be divisible by 3. That means the number of 2s multiplied by 2 must be a multiple of 3. Therefore, the number of 2s must be a multiple of 3 (since 2 and 3 are coprime). So, the number can't have 1 or 2 twos; it needs at least 3 twos.

3. Divisible by 5: The last digit must be 0 or 5. But since we can only use 0 and 2, the last digit must be 0. So, the number must end with a 0. That means the number has to be a multiple of 10, so the last digit is 0.

Putting these together, the number must end with 0, have a sum of digits divisible by 3 (so at least three 2s), and be as small as possible. So, the number will be something like 222...220, where the total number of 2s is a multiple of 3, and the last digit is 0. We need to find the smallest such number.

Let me start by checking the smallest possible numbers that fit these criteria.

First, the number must end with 0. So, the smallest possible number is 20. Let's check divisibility by 3: 2 + 0 = 2, which is not divisible by 3. So, 20 is not a multiple of 30. Next, try 220. The sum is 2 + 2 + 0 = 4, still not divisible by 3. Next, 2220. Sum is 2 + 2 + 2 + 0 = 6, which is divisible by 3. So, 2220. Let's check if it's divisible by 5: yes, ends with 0. Divisible by 2: yes. Divisible by 3: yes. So, 2220 is a multiple of 30. Is that the smallest?

Wait, but maybe there's a smaller number with fewer digits. Let's see. Since the number must end with 0, the smallest possible number would be 20, 220, 2220, etc. Since 20 is not divisible by 3, the next candidate is 220, which is 220. Let's check 220: 2+2+0=4, not divisible by 3. So, 220 isn't. Then 2220: sum 6, which is divisible by 3. So, 2220 is the next candidate. Is there a number with 3 digits? Let's see. If the number is three digits, it must end with 0. So, possible numbers are 220 and 200. Wait, 200 is also a three-digit number. Let's check 200: sum is 2 + 0 + 0 = 2, not divisible by 3. 220: sum 4, not divisible by 3. So, no three-digit numbers work. Then four digits: 2220 is 2220. Let's check if there's a four-digit number smaller than 2220. Wait, but 2220 is the first four-digit number. Is there a smaller four-digit number that ends with 0 and has three 2s? Hmm. Let's see. The four-digit numbers ending with 0 are 2220, 2200, 2020, 2000. Let's check those:

- 2220: sum 6, divisible by 3. So, yes.
- 2200: sum 2 + 2 + 0 + 0 = 4, not divisible by 3.
- 2020: sum 2 + 0 + 2 + 0 = 4, same issue.
- 2000: sum 2, same.

So, 2220 is the only four-digit number ending with 0 with three 2s. Therefore, 2220 is the candidate. But wait, maybe there's a five-digit number that's smaller than 2220? Wait, no, 2220 is a four-digit number. Five-digit numbers would be larger, so 2220 is the smallest. But let me confirm.

Wait, 2220 is 2220. But maybe a three-digit number with two 2s and one 0? Wait, but three-digit numbers ending with 0 can only have two digits before. Let's see: 220, 200, 220, 200. Wait, we already checked those. They don't have a digit sum divisible by 3.

Alternatively, maybe a four-digit number with three 2s and one 0. The smallest four-digit number would be 2020, but sum is 4. Then 2200, sum is 4. Then 2220, sum 6. So, 2220 is the first one that works. Therefore, 2220 is the answer.

But wait, let me check if there's a smaller number. Wait, 2220 is 2220, which is 2220. Let me see if there's a two-digit number: 20, sum 2. No. So, three-digit numbers: 200, 220, 202. Wait, 202 is 202, sum 4. 200: sum 2. 220: sum 4. 222 is three digits, but ends with 2, so not allowed. So, no three-digit numbers. Then four-digit numbers: 2220 is the first one that works.

But wait, maybe 2020? Wait, 2020 ends with 0, sum is 2+0+2+0=4. Not divisible by 3. 2200: sum 4. 2000: sum 2. 2220: sum 6. So, 2220 is the first one. Therefore, 2220 is the answer.

But let me check another angle. Maybe there's a number with more digits but smaller in value? For example, 20020. Wait, but that's five digits, which is larger than 2220. So, 2220 is smaller. Alternatively, 2220 is 2220, which is 2,220. Let me see if there's a number between 200 and 2220 that works. For example, 2020 is 2020, which is 2,020. Sum is 4. Not divisible by 3. Then 2002: sum 4. 2022: ends with 2, invalid. 2202: ends with 2, invalid. 2220: works. So, yes, 2220 is the smallest.

But let me verify once more. Let's think of numbers ending with 0 and with digit sums divisible by 3. The digit sum must be a multiple of 3, but since the digits are 2s and 0s, the number of 2s must be a multiple of 3. So, the minimal number of 2s is 3. Therefore, the number must have at least three 2s and end with 0. So, the minimal number is 2220. Let's check 2220 divided by 30: 2220 / 30 = 74. So, yes, it's an integer. Therefore, 2220 is the correct answer.

But wait, is there a number with more digits but smaller in value? For example, 20020, but that's five digits and 20020 is larger than 2220. So, no. Therefore, 2220 is the smallest. Therefore, the answer is 2220.
Therefore, the least positive integer multiple of $30$ that can be written with only the digits $0$ and $2$ is $\boxed{2220}$.
</think>